Problem 46 - Contemporary Precalculus

Question

Find the equation of the ellipse that satisfies the given conditions. Center (-5,2)\(;\) endpoints of major and minor axes: (0,2), (-5,17),(-10,2),(-5,-13).

Step-by-Step Solution

Answer

Answer: The equation of the ellipse is \(\frac{(x+5)^2}{225}+\frac{(y-2)^2}{25}=1\).

Step 1: Determine the center

We are given the center of the ellipse as (-5, 2), so (h, k) = (-5, 2).

Step 2: Determine the length of the major and minor axes

We are given the endpoints of the major and minor axes as (0,2), (-5,17), (-10,2), and (-5,-13). By finding the distance between the center and these points, we can determine the length of the major and minor axes. The distance formula is given by \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). For point (0,2), the distance to the center (-5,2) is: \(d = \sqrt{(-5-0)^2+(2-2)^2} = \sqrt{(-5)^2} = 5\) For point (-5, 17), the distance to the center (-5,2) is: \(d = \sqrt{(-5-(-5))^2+(17-2)^2} = \sqrt{(0)^2+(15)^2} = 15\) Since the major axis is the longest of the two axes, the length of the major axis (a) is 15, and the length of the minor axis (b) is 5.

Step 3: Write the equation of the ellipse

Now that we have the center (h,k) = (-5, 2), the length of the major axis a = 15, and the length of the minor axis b = 5, we can write the equation of the ellipse using the standard form: \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\) \(\frac{(x-(-5))^2}{15^2}+\frac{(y-2)^2}{5^2}=1\) \(\frac{(x+5)^2}{225}+\frac{(y-2)^2}{25}=1\) This is the equation of the ellipse that satisfies the given conditions.

Key Concepts

Center of an Ellipse

The center of an ellipse is a crucial point that defines the location of the ellipse in the coordinate plane. It is represented by the coordinates \(h, k\).
For an ellipse given by the standard equation \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), the center of the ellipse is always at \(h, k\).
In our exercise, the center is provided as \((-5, 2)\), which means every calculation for the distance will revolve around this point. It serves as a reference point from which you can determine the positions of the vertices on the ellipse.
The center helps in constructing the equations required for the ellipse and guides the determination of the major and minor axes.

Distance Formula

The distance formula is used to calculate the length between two points in a plane. It is essential in understanding and determining the dimensions of the ellipse.
The general form of the distance formula is \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), where \( (x_1, y_1) \) and \( (x_2, y_2) \) are two points in the plane.
In the ellipse problem, we utilize this formula to find the distance from the center to each of the endpoints of the axes.

• For point \((0, 2)\): The distance is 5, indicating a minor axis endpoint.
• For point \((-5, 17)\): The distance is 15, identifying it as a major axis endpoint.

This calculation is key to properly defining the lengths of both axes, necessary for forming the equation of the ellipse.

Major Axis

The major axis is the longest diameter of the ellipse, stretching from one end of the ellipse to the other, passing through the center.
The endpoints of the major axis lie on these coordinates: \((-5, 17)\) and \((-5, -13)\). This vertical orientation tells us that the ellipse is elongated vertically.
The length of the major axis is determined by calculating the distance from the center to these endpoints. In our solution, the measure of the major axis is found using:

• Distance from the center \((-5,2)\) to \((-5, 17)\): 15 units.

Therefore, the value of the semi-major axis is \(a = 15\), meaning the total length of the major axis is twice that number, making it 30 units long, centered at \((-5, 2)\).

Minor Axis

In contrast, the minor axis is the shorter diameter of the ellipse. It is perpendicular to the major axis and also passes through the center.
The endpoints of the minor axis are positioned at \( (0, 2) \) and \((-10, 2)\), showing that the ellipse is narrow horizontally.
Using the distance formula once more, we find the length of the semi-minor axis. Starting from the center \( (-5, 2)\) to endpoint \( (0, 2)\), the distance is calculated as:

• 5 units which indicates the total length of the minor axis as 10 units (double the semi-minor length).

This allows the complete formation of the ellipse along both axes, setting the parameters for accurately describing the ellipse's dimensions in the coordinate system.